Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{t^2 - 6t}{t^2 - 16t + 60} \div \dfrac{9t - 63}{t - 10} $
Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{t^2 - 6t}{t^2 - 16t + 60} \times \dfrac{t - 10}{9t - 63} $ First factor the quadratic. $a = \dfrac{t^2 - 6t}{(t - 10)(t - 6)} \times \dfrac{t - 10}{9t - 63} $ Then factor out any other terms. $a = \dfrac{t(t - 6)}{(t - 10)(t - 6)} \times \dfrac{t - 10}{9(t - 7)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ t(t - 6) \times (t - 10) } { (t - 10)(t - 6) \times 9(t - 7) } $ $a = \dfrac{ t(t - 6)(t - 10)}{ 9(t - 10)(t - 6)(t - 7)} $ Notice that $(t - 6)$ and $(t - 10)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ t(t - 6)\cancel{(t - 10)}}{ 9\cancel{(t - 10)}(t - 6)(t - 7)} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $a = \dfrac{ t\cancel{(t - 6)}\cancel{(t - 10)}}{ 9\cancel{(t - 10)}\cancel{(t - 6)}(t - 7)} $ We are dividing by $t - 6$ , so $t - 6 \neq 0$ Therefore, $t \neq 6$ $a = \dfrac{t}{9(t - 7)} ; \space t \neq 10 ; \space t \neq 6 $